Three Phase Controlled Rectifiers2003

8/13/2019 Three Phase Controlled Rectifiers2003

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VAN

VCN

VBN

1200

1200

1200 RN AN

YN BN

BN CN

v v

v v

v v


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T1

T2

T3

Rload

3-supply


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PhaseVoltages

0 30 60 90 120 150 180 210 240 270 300 330 360

Vo(=0 )

Vo(30+

)

Va Vb Vc

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360


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PhaseVoltages

0 30 60 90 120 150 180 210 240 270 300 330 360

Vo(=0 )

Vo(30+

)

Va Vb Vc

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360


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01

0

2

0

3

0

306

5

1506

7 270

62

Each thytistor conducts for 120 or radians3

T is triggered at t

T is triggered at t

T is triggered at t


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5

6

6

If the reference phase voltage is

sin , the average or dc output

voltage for continuous load current is calculated

using the equation

3

sin .2

RN an m

dc m

v v V t

V V t d t


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5

6

6

56

6

3 sin .2

3cos

2

3 5cos cos

2 6 6

mdc

mdc

mdc

VV t d t

VV t

VV


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0 0

0

Note from the trigonometric relationship

cos cos .cos sin .sin

5 5cos cos sin sin

6 63

2co

cos 150 cos sin 150 sin32 cos 30

s .cos sin sin6 6

.cos

mdc

m

dc

A

VV

B A B A B

V

V

0sin 30 sin


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0 0

0 0 0 0

0 0

0 0

0

0

0

0

0 0

Note: cos 1

cos 180 30 cos sin 180 30 sin3

2 cos 30 .cos sin 30 sin

cos 30 cos sin 30 sin3

2 cos 30 .cos sin 30 s

80 30 cos 30

sin 180 30 sin 30

in

m

dc

m

dc

VV

V

V


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03 2cos 30 cos2

3 32 cos

2 2

3 3 33 cos cos

2 2

3 cos2

Where 3 Max. line to line supply voltage

m

dc

m

dc

m m

dc

Lm

dc

Lm m

VV

VV

V VV

VV

V V


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15 26

2 2

6

12

The rms value of output voltage is found by

using the equation

3sin .

2

and we obtain

1 33 cos 2

6 8

mO RMS

mO RMS

V V t d t

V V


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R-Lload

T1 T3 T5

T4 T2T6

A

B

C

6 - pulse combinations are:

(6,1) (1,2) (2,3) (3,4) (4,5) (5,6)

Io


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R-Lload

T1 T3 T5

T4 T6

A

B

C

When thyristors T1& T6conducts

T2

A T1 RL T6 B

Io


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R-Lload

T1 T3 T5

T4 T6

A

B

C


T2

A T1 RL T2 C

Io


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R-Lload

T1 T3 T5

T4 T6

A

B

C


T2

B T3 RL T2 C

Io


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R-Lload

T1 T3 T5

T4 T6

A

B

C


T2

B T3 RL T4 A

Io


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R-Lload

T1 T3 T5

T4 T6

A

B

C


T2

C T5 RL T4 A

Io


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R-Lload

T1 T3 T5

T4 T6

A

B

C


T2

C T5 RL T6 B

Io


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Line

Voltages

PhaseVoltages

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360


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LineVoltages

Vo

(=0 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360VCBVCAVAB VBAVBCVAC

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360Vo(=60 )

Io


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LineVoltages

Vo

(=0 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360VCBVCAVAB VBAVBCVAC

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360

Vo

(=60 )

Io


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0 30 60 90 120 150 180 210 240 270 300 330 360

LineVoltages

Vo

(=90 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360

Vo

(=120 )0 30 60 90 120 150 180 210 240 270 300 330 360


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0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360

LineVoltages

Vo

(=90 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360

Vo

(=120 )


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The output load voltage consists of 6 voltage

pulses over a period of 2radians, Hence the

average output voltage is calculated as

2

6

6 . ;2

3 sin6

dc OO dc

O ab m

V V v d t

v v V t


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2

6

mL

max

3

3 sin .6

3 3 3cos cos

Where V 3 Max. line-to-line supply vo

The maximum average dc output voltage is

obtained for a delay angle

ltage

3 3 0,

3

dc m

m mL

dc

m

m m

dmdc

V V t d t

V VV

V

V VV V

L


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1

22

2

6

The normalized average dc output voltage is

cos

The rms value of the output voltage is found from

6.2

dc

dcn n

dm

OO rms

VV V

V

V v d t


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1

22

2

6

1

22

2 2

6

1

2

6 .2

33 sin .

2 6

1 3 33 cos 2

2 4

abO rms

mO rms

mO rms

V v d t

V V t d t

V V


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For four quadrant operation in many industrial variable speed dcdrives , 3 phase dual converters are used.

Used for applications up to 2 mega watt output power level.

Dual converter consists of two 3 phase full converters which areconnected in parallel & in opposite directions across a common load.

+I0-I0

+V0

-V0

Conv. 1

Inverting

1 > 900

Conv. 1

Rectifying

1 < 900

Conv. 2

Rectifying

2 < 900

Conv. 2Inverting

2 > 900


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The Dual Converter is operated such that:

One conv. should be Converter One conv. should be Inverter

{1 < 90}{2 > 90}

i.e, 1+2= 180

Conditions:

1. O/P voltage mag. of 2 converters should be same.

2. Polarities should be same.

i.e,Vo = Vo1 = -Vo2


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Non Circulating current mode:

only one converter is in operation at a time, and it alonecarries the entire load current.

No Reactor is used between two converters.

Circulating current mode:

A reactor is inserted between two reactors. Thisreactor limits the magnitude of circulating current

to a reasonable value.


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load

V0

T11 T13 T15

T14 T12T16 T21T23T25

T24T22 T26

A

B

C C

B

A

V01 V02

Lr2

Lr2

Vr Vr

2 2


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LineVoltages

Vo1

(=60 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360VCB V

CA

VAB

VBA

VBC

VAC

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360Vo2(=120 )

Io


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LineVoltages

Vo1

(=60 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360VCB V

CA

VAB

VBA

VBC

VAC

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360Vo2(=120 )

Vo=Vo1+Vo2

Vo2 (=120 )

2


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LineVoltages

Vr

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360

Ic =

Vr dt

Ic


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1 2

3 sin sin6 2

3 cos6

The circulating current can be calculated by

using the equation

r O O ab bc

r m

r m

v v v v v

v V t t

v V t


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1

1

6

6

1

max

1.

13 cos .

6

3sin sin

63

t

r r

r

t

r m

r

m

r

r

m

r

r

i t v d t

L

i t V t d t L

Vi t t

L

Vi

L


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Because of Source Inductance effect, overlap conductionof thyristors takes place.

At an instant, 3 thyristors will conduct

i.e, 2 thyristors from +ve group& 1 thyristor from -ve group

(or)

2 thyristors from -ve group& 1 thyristor from +ve group

So, .output voltage decreases.


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Decrease in output voltage due to overlap angle:

Vo = Area of the triangle

Vo = 6x (1/2)0Vl d(t)VO = 3Ls

Io

Io

For 3- full bridge conv. With R-L Load without Ls

VO= 33 Vm cos

Therefore, Avg. o/p voltage with overlap angle (or) Source Inductance

VO(avg)= 33 Vm cos

3Ls

Io


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3 Phase semiconverters are used in Industrial dc driveapplications upto 120kW power output.

Single quadrant operation is possible.

Power factor decreases as the delay angle increases.

Power factor is better than that of 3 phase half waveconverter.


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R-Lload

T1 T3 T5

D4 D 2D6

A

B

C

Io


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R-Lload

T1 T3 T5

D4 D6

A

B

C


D2

A T1 RL T6 B

Io


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LineVoltages

Vo

(=30 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360

VCBVCAVAB VBAVBCVAC

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360Vo(=60 )

Io


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LineVoltages

Vo

(=30 )

AB AC BC BA CA CB

0 30 60 90 120 150 180 210 240 270 300 330 360

VCBVCAVAB VBAVBCVAC

0 30 60 90 120 150 180 210 240 270 300 330 360

0 30 60 90 120 150 180 210 240 270 300 330 360

Vo(=60 )

IphA

1,2 3,4 5,6

Phase A For Inductive LOAD


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To derive an Expression for the

Average Output Voltage of 3 PhaseSemiconverter for > / 3and Discontinuous Output Voltage


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76

6

76

6

For and discontinuous output voltage:

3the Average output voltage is found from

3 .2

33 sin

2 6

dc ac

dc m

V v d t

V V t d t


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max

3 31 cos

23

1 cos2

3 Max. value of line-to-line supply voltage

The maximum average output voltage that occurs at

a delay angle of 0 is

3 3

m

dc

mL

dc

mL m

m

dmdc

VV

VV

V V

VV V


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17 2

6

2

6

The normalized average output voltage is

0.5 1 cos

The rms output voltage is found from

3.

2

dcn

dm

acO rms

VV

V

V v d t


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Three Phase Controlled Rectifiers2003

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